diff --git a/archive/hugo/content/resources.md b/archive/hugo/content/resources.md
index 35cba6e..00e05f6 100644
--- a/archive/hugo/content/resources.md
+++ b/archive/hugo/content/resources.md
@@ -52,6 +52,8 @@ Recently, I've been...
 
 | Name | Progress  |
 | --- | --- |
+| [獵魔士](https://www.netflix.com/hk/title/80189685) |
+| [](https://baike.baidu.com/item/罗小黑战记/22902442) |
 | [罗小黑战记](https://baike.baidu.com/item/罗小黑战记/22902442) |
 | [哪吒之魔童降世](https://movie.douban.com/subject/26794435//) |
 | [中国机长](https://movie.douban.com/subject/30295905/) |
diff --git a/archive/hugo/content/tools.md b/archive/hugo/content/tools.md
index f70f1ec..415260b 100644
--- a/archive/hugo/content/tools.md
+++ b/archive/hugo/content/tools.md
@@ -34,3 +34,9 @@ menu = "main"
 ### iOS
 
 - [Infuse](https://firecore.com/infuse)：视频播放
+
+## 网站
+
+### 下载
+
+- [Old version](http://www.oldversion.com)：下载软件历史版本
\ No newline at end of file
diff --git a/source/_posts/daily-trivials.md b/source/_posts/daily-trivials.md
index 0f8208d..2158400 100644
--- a/source/_posts/daily-trivials.md
+++ b/source/_posts/daily-trivials.md
@@ -8,6 +8,12 @@ draft: false
 
 这篇文章讲用于记录我日常遇到的一些问题的解决和一些小的hacks。话不多说，let's begin.
 
+
+- start zookeeper error: ERROR Unexpected exception, exiting abnormally (org.apache.zookeeper.server.ZooKeeperServerMain)
+java.io.IOException: No snapshot found, but there are log entries. Something is broken!
+
+手动删除掉zookeeper下的数据后可以正常启动 `rm -rf /usr/local/var/lib/zookeeper`
+
 - nginx测试配置是否正确
 
 The -c flag indicates a certain configuration file will follow; the -t flag tells Nginx to test our configuration. 
@@ -38,9 +44,9 @@ chmod 777 /tmp/.X11-unix/X1
    brew install kafka
 ```
 	To have launchd start kafka now and restart at login:
-	`brew services start kafka`
+	brew services start kafka
 	Or, if you don’t want/need a background service you can just run:
-	`zookeeper-server-start /usr/local/etc/kafka/zookeeper.properties && kafka-server-start /usr/local/etc/kafka/server.properties`
+	zookeeper-server-start /usr/local/etc/kafka/zookeeper.properties && kafka-server-start /usr/local/etc/kafka/server.properties
 
 - 误删macOS唯一一个administrator的用户admin组，不小心使用了`sudo dseditgroup -o edit -a $(whoami) -t user admin` （本意只是想从wheel组里面删掉）命令把系统上唯一一个具有超级管理员权限的用户的组权限给删了，导致使用sudo命令会提示 `$(user) not in sudoers file, this incident will be reported....`
 
diff --git a/source/_posts/leetcode-7-reverse-integer-md.md b/source/_posts/leetcode-7-reverse-integer-md.md
new file mode 100644
index 0000000..1e4f10c
--- /dev/null
+++ b/source/_posts/leetcode-7-reverse-integer-md.md
@@ -0,0 +1,67 @@
+---
+title: leetcode-7-reverse-integer.md
+tags: ["leetcode", "solution"]
+date: 2019-11-26 20:56:46
+---
+
+	7. Reverse Integer
+	Given a 32-bit signed integer, reverse digits of an integer.
+	Example 1:
+	
+	Input: 123
+	Output: 321
+	Example 2:
+	
+	Input: -123
+	Output: -321
+	Example 3:
+	
+	Input: 120
+	Output: 21
+	
+	Note:
+	Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231,  231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
+
+<!--more-->
+
+题意很简单，就是数字反转，例如123-\>321。解法也比较简单，就是每次取除10的模数（然后除10取整），原数乘10累加。例如123就是1+10*((0+3)*10+2)=321。需要注意有两个点，一个是可能数字是120000，那么就不能简单地通过字符串反转来做（算数方法做是没问题的）；另一个就是有边界，边界是4字节的有符号整型数字的范围，也就是 $$[-2^{-31}, 2^{31}-1]$$。
+
+我的做法比较粗暴，判断还有一位的时候用边界去减了除10，和当前数比较大小即可（其实这么做有问题，如果test case长度不是2的31次方位，比如100位的数字，那也会最后一步才会判断是否溢出，对于python这种科学计算型语言不会有问题，因为底层已经处理了大数运算。对于C、Java等传统语言，这么写就会有问题。当然，也可以上来就手动判一下数字长度，如果大于10就直接返回0，不过我没写。），也没想到居然这么写过了，代码如：
+
+```python
+class Solution:
+    def reverse(self, x: int) -> int:
+        n = 0
+        flag = 1
+        if x < 0:
+            flag = -1
+            x = -x
+        while x > 0:
+            if x < 10:
+                if flag == 1 and n > ((1<<31)-1-x)/10 or flag == -1 and n > ((1<<31)-1)/10:
+                    return 0
+            n = n * 10 + x % 10
+            x = x // 10
+        return flag * n
+```
+
+
+
+然后就是参考题解了。发现有两个地方可以改进代码。一个其实不用一开始就把负数转换为正数，因为带符号数累加之后符号还是在的；还有一个是判断条件可以做修改为判断为当前数是不是大于MAXINT/10，因为如果大于的话那么下一次x10就必然溢出；如果等于那么判断尾数是不是大于等于8（正数时）、小于等于-9（负数时）。
+
+
+
+
+
+
+
+
+
+
+# Refer
+
+[7. Reverse Integer](https://leetcode.com/problems/reverse-integer/)
+
+[7. Reverse Integer](https://leetcode.com/articles/reverse-integer/)
+
+# Appendix
\ No newline at end of file